Tip: \(A_{parallelogram} = bh,\) where \(b\) is its base, and \(h\) is its height.Tip: Area of a triangle with height \(h\) and base \(b\): \(A_{\triangle} = \frac{1}{2}bh.\)
To find the area of the parallelogram, we must know the lengths of its base and height. The length of its base is \(AB = AE + EB = 5+ 15 = 20,\) and we can find the length of its height, \(DE,\) by using the information about \(\triangle AED.\)

(A)Tip: \(A_{parallelogram} = bh,\) where \(b\) is its base, and \(h\) is its height.
If you use \(A_{\triangle} = \frac{1}{2}bh\) to find the area of the parallelogram, you will get this wrong answer.

(B)
If you estimate that about 5 triangles with the same area as \(\triangle AED\) would fit into the parallelogram, you may get this wrong answer. Notice that the diagram is not drawn to scale.

(C)Tip: Read diagrams carefully.
If you think the base of the parallelogram is 15, you will get this wrong answer. The base is \(\overline{AB}\) and \(AB = AE + EB = 5+ 15 = 20.\)

(E)Tip: \(A_{parallelogram} = bh,\) where \(b\) is its base, and \(h\) is its height.
If you find the area of the parallelogram by multiplying \(DA\) and \(AB,\) you will get this wrong answer. \(DA\) is not the parallelogram's height.

In the figure above, the four small squares are congruent. What is the area of the shaded region?

Other Polygons

Tip: The sum of the measures of the exterior angles, one per vertex, of any convex polygon is \(360^\circ.\)
If we count one exterior angle per vertex, then the equilateral triangle has 3 congruent exterior angles and the regular octagon has 8 congruent exterior angles. So,